Explanation

(c)
Given equation of hyperbola is $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ and equation of conjugate hyperbola is $\frac{x^{2}}{b^{2}} - \frac{y^{2}}{a^{2}} = 1.$
Since, $e$ and $e'$ are the eccentricities of the respective hyperbola, then
$e^{2} = 1 + \frac{b^{2}}{a^{2}},\ \left( e^{'} \right)^{2} = 1 + \frac{a^{2}}{b^{2}}$
$\therefore\ \frac{1}{e^{2}} + \frac{1}{e^{'2}} = \frac{a^{2}}{a^{2} + b^{2}} + \frac{b^{2}}{a^{2} + b^{2}}.$